Question

我的代码有问题，但我知道问题是什么。等级有2个字符，因此显示2个结果，每个结果加0.3。这是一个问题，因为只有一个结果应该打印到控制台。

keydict = {'A': 4.0,'B': 3.0,'C': 2.0,'D': 1.0,'F': 0, '+': 0.3, '-': -0.3}
grade = input('Grade:')


def letter2number(letter):
    if len(grade) > 2:
        print('Too many characters')
        letter2number(grade)
    for char in grade:
        if char in keydict:
            if '+' in grade:
                print(keydict[char] + keydict['+'])
            elif '-' in grade:
                print(keydict[char] + keydict['-'])
            else:
                print(keydict[char])
        else:
            print(grade,'is an invalid input')
    return


letter2number(grade)

这是在用户输入＆＃34; A +＆＃34;（或任何等级为＆＃39; +＆＃39;或＆＃39; - ＆＃39;我意识到for循环导致了这个问题，因为输入中有2个字符。但是，如果不使用for循环，我认为还有另一种方法可以做到这一点。预计4.3，0.6不是我想要的。

Grade: A+ # <<< Thats a user input
4.3       # <<< YES
0.6       # <<< How can I get this to NOT print

除了正在打印的0.6之外，我对代码没有其他问题。换句话说，我不希望0.6在那里，但4.3应该留下来。

Answer 1

此处不需要for循环，导致您的问题。请记住，for循环将独立查看两个字符

"A"  # evaluates to 4.0, add 0.3 since '+' exists in the string
"+"  # evaluates to 0.3, add 0.3 since '+' exists in the string

这绝对不是你想要的。你应该做的是：

def lettertograde(letter):
    letter, *modifier = letter
    grade = keydict[letter]
    if "+" in modifier:
        grade += 0.3
    elif "-" in modifier:
        grade -= 0.3
    return grade

这个a, *b = ...语法是splat解包，并通过从迭代中拉出部分来实现：

a, *b = "A"          # a = "A", b = []
a, *b = "A+"         # a = "A", b = ["+"]
a, *b = "A-"         # a = "A", b = ["-"]
a, *b = "Good job!"  # a = "G", b = ["o", "o", "d", " ", "j", "o", "b", "!"]
a, *b = ""           # ValueError("not enough values to unpack (expected at least 1, got 0)")

N.B。同一表达式中的多个splats引入歧义，并且是不允许的。

a, *b, c, *d = "anything"  # SyntaxError("two starred expressions in assignment")

Answer 2

>>> keydict = {'A': 4.0,'B': 3.0,'C': 2.0,'D': 1.0,'F': 0, '+': 0.3, '-':    0.3, ' ': 0.0}
>>> grade = input('Grade:') + ' '
>>> if grade[0] in keydict and grade[1] in keydict:
       print(grade, keydict[grade[0]]+ keydict[grade[1]])
    else:
       print(grade, 'is an invalid input')

B 3.0 香蕉是无效的输入 B-2.7

Answer 3

您需要在块break

的末尾添加if char in keydict:

keydict = {'A': 4.0,'B': 3.0,'C': 2.0,'D': 1.0,'F': 0, '+': 0.3, '-': -0.3}
grade = input('Grade:')


def letter2number(letter):
    if len(grade) > 2:
        print('Too many characters')
        letter2number(grade)
    for char in grade:
        if char in keydict:
            if '+' in grade:
                print(keydict[char] + keydict['+'])
            elif '-' in grade:
                print(keydict[char] + keydict['-'])
            else:
                print(keydict[char])

            break  # <<---

        else:
            print(grade,'is an invalid input')
    return


letter2number(grade)

break停止循环

Answer 4

生成修改后的成绩并将其添加到词典中。然后查看整个成绩字符串。

>>> keydict = {'A': 4.0, 'B': 3.0, 'C': 2.0, 'D': 1.0, 'F': 0}
>>> modifiers = {'+': 0.3, '-': -0.3}
>>> keydict.update({ k + mk: v + mv for k, v in keydict.items() for mk, mv in modifiers.items() })
>>> keydict
{'D': 1.0, 'F+': 0.3, 'A+': 4.3, 'F': 0, 'F-': -0.3, 'C+': 2.3, 'B': 3.0, 'C': 2.0, 'D-': 0.7, 'C-': 1.7, 'B-': 2.7, 'D+': 1.3, 'B+': 3.3, 'A-': 3.7, 'A': 4.0}
>>> grade = 'C-'
>>> keydict[grade]
1.7

如何只将1个结果打印到控制台而不是两个都打印出来？

4 个答案: